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Midterm solutions

# Midterm solutions - ECE 135 MIDTERM SOLUTION 1 See class...

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ECE 135 MIDTERM SOLUTION 1. See class notes. 2. (a) Broadening factor of a Gaussian pulse of initial rms width 0 σ is given as ( ) ( ) 2 2 2 2 2 2 2 3 2 2 2 2 2 3 0 0 0 0 1 1 1 2 2 4 2 L C L L V C V ω ω β β β σ σ σ σ σ = + + + + + + , where 0 2 V ω ω σ σ = . ω σ is the rms spectral width of the source. Source spectrum is assumed to be Gaussian. C is the chirp parameter. If there is no chirp and the source spectral width is very narrow in the presence of dispersion only (i.e., when dispersion slope is negligible) this equation simplifies to 2 2 2 2 2 0 0 1 2 L β σ σ σ = + or 2 2 2 2 0 0 2 L β σ σ σ = + or 2 2 2 0 0 2 L β σ σ σ = + . Obviously there is an optimum 0 σ to minimize σ or the broadening. This can be found by differentiating the above equation and setting 0 0 d d σ σ = . This yields 2 2 0 2 0 0 2 0 2 2 0 0 2 2 2 2 1 0 2 2 L L d d L β β σ σ σ σ σ β σ σ + = = + Hence 2 2 0 3 0 1 0 2 L β σ σ = 2 4 2 0 2 L β σ = 2 0 2 L β σ = . If we substitute this optimum 0 σ value into the broadening equation we get 2 2 2 2 2 2 2 2 L L L L β β σ β β = + = . The maximum bit rate is found using the condition 4 1 B σ as 2 1 4 B L β .

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(b) 2 β is found using 2 2 2 c D π β λ = − as 2 2 2 D c λ β π = − . ( ) 2 6 2 12 2 12 2 26 2 3 9 8 8 1.5 10 m ps 2.25 10 m 10 s s 15 15 1.79 10 m m km-nm 10 m -10 m m 2 3 10 6 10 s s β π π × × = − × = − × = − × × × × 2 24 2 2 26 26 2 3 s 10 ps ps 1.79 10 1.79 10 17.9 m 10 km km β = − × = − × = − Hence 2 2 0 ps 1 17.9 100 km 29.9 ps 2 km 2 L β σ = = × × = and 3 12 2 2 1 1 1 5.9 10 1 GBits 0.0059 5.9 ps 10 s s 4 ps 4 17.9 100 km km B L β × = = = = × (c) If there is chirp broadening factor becomes 2 2 2 2 2 2 2 2 0 0 0 1 2 2 C L L β β σ σ σ σ = + + or 2 2 2 2 2 0 0 0 2 2 C L L β β σ σ σ σ = + + or 2 2 2 2 0 0 0 2 2 C L L β β σ σ σ σ
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Midterm solutions - ECE 135 MIDTERM SOLUTION 1 See class...

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